Problem 1705

Solution:
Assume throughout that $a, b, c$ are nonnegative integers. Since the prime factorization of $2004$ is $2004 = 2^{2} \cdot 3 \cdot 167$,
$$T = \left\{ 2^{a} 3^{b} 167^{c} \mid 0 \leq a \leq 200, 0 \leq b, c \leq 100 \right\}$$
Let
$$S = \left\{ 2^{200-b-c} 3^{b} 167^{c} \mid 0 \leq b, c \leq 100 \right\}$$
For any $0 \leq b, c \leq 100$, we have $0 \leq 200-b-c \leq 200$, so $S$ is a subset of $T$. Since there are $101$ possible values for $b$ and $101$ possible values for $c$, $S$ contains $101^{2}$ elements. We will show that no element of $S$ is a multiple of another and that no larger subset of $T$ satisfies this condition.

Suppose $2^{200-b-c} 3^{b} 167^{c}$ is an integer multiple of $2^{200-j-k} 3^{j} 167^{k}$. Then
$$200-b-c \geq 200-j-k, \quad b \geq j, \quad c \geq k$$
But this first inequality implies $b+c \leq j+k$, which together with $b \geq j, c \geq k$ gives $b = j$ and $c = k$. Hence no element of $S$ is an integer multiple of another element of $S$.

Let $U$ be a subset of $T$ with more than $101^{2}$ elements. Since there are only $101^{2}$ distinct pairs $(b, c)$ with $0 \leq b, c \leq 100$, then (by the pigeonhole principle) $U$ must contain two elements $u_{1} = 2^{a_{1}} 3^{b_{1}} 167^{c_{1}}$ and $u_{2} = 2^{a_{2}} 3^{b_{2}} 167^{c_{2}}$, with $b_{1} = b_{2}$ and $c_{1} = c_{2}$, but $a_{1} \neq a_{2}$. If $a_{1} > a_{2}$, then $u_{1}$ is a multiple of $u_{2}$ and if $a_{1} < a_{2}$, then $u_{2}$ is a multiple of $u_{1}$. Hence $U$ does not satisfy the desired condition.

Therefore the largest possible number of elements that such a subset of $T$ can have is $101^{2} = 10201$.